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Unformatted text preview: MATH 110: LINEAR ALGEBRA SPRING 2007/08 PROBLEM SET 8 SOLUTIONS 1. Let V be a vector space over F and S , T End( V ). (a) Show that I  S T is injective iff I  T S is injective. Solution. The trick is to observe that S ( I  T S ) = ( I  S T ) S . Suppose I S T is injective. Let v ker( I T S ). Then ( I T S )( v ) = V . Hence ( I  S T )( S ( v )) = S (( I  T S )( v )) = S ( V ) = V and since I  S T is injective, we deduce that S ( v ) = V . So T S ( v ) = T ( V ) = V ; and so v = v V = v T S ( v ) = ( I  T S )( v ) = V where the last equality follows from our choice of v ker( IT S ). Hence ker( IT S ) = { V } and we deduce that I  T S is injective by Theorem 4.12 . (b) T is called nilpotent if T n = O for some n N . Show that if T is nilpotent, then I  T is bijective. What is ( I  T ) 1 ? Solution. Let S = I + T + T 2 + + T n 1 (by the geometric series heuristic in lecture). Then ( I  T ) S = ( I + T + + T n 1 ) ( T + T 2 + + T n ) = I  T n = I  O = I and likewise for S ( I  T ). Hence ( I  T ) 1 = S and I  T is bijective. 2. Let V be a vector space over R and T End( V ) be an involution, ie. T 2 = I . Define V + := { v V  T ( v ) = v } and V := { v V  T ( v ) = v } . (a) Show that V + and V are subspaces of V . Solution. Let 1 , 2 R and v 1 , v 2 V + . Then T ( v 1 ) = v 1 , T ( v 2 ) = v 2 , and T ( 1 v 1 + 2 v 2 ) = 1 T ( v 1 ) + 2 T ( v 2 ) = 1 v 1 + 2 v 2 . Hence 1 v 1 + 2 v 2 V + . Likewise for V . Date : May 7, 2008 (Version 1.0). 1 (b) Show that V + V = V. Solution. Note that v = 1 2 ( I + T )( v ) + 1 2 ( I  T )( v ) . (2.1) Since T 2 = I , T ( 1 2 ( I + T )( v ) ) = T ( 1 2 v + 1 2 T ( v ) ) = 1 2 T ( v ) + 1 2 T 2 ( v ) = 1 2 T ( v ) + 1 2 I ( v ) = 1 2 ( I + T )( v ) and likewise, T ( 1 2 ( I  T )( v ) ) = T ( 1 2 v 1 2 T ( v ) ) = 1 2 T ( v ) 1 2 T 2 ( v ) = 1 2 T ( v ) 1 2 I ( v ) = 1 2 ( I  T )( v ) ....
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 Spring '08
 GUREVITCH
 Math, Linear Algebra, Algebra, Vector Space

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